Answer
(a) $xy$-plane
$\mathop \smallint \limits_{}^{} \mathop \smallint \limits_{}^{} \mathop \smallint \limits_{\cal W}^{} z{\rm{d}}V = \mathop \smallint \limits_{x = 0}^1 \mathop \smallint \limits_{y = 2x}^2 \mathop \smallint \limits_{z = 0}^{2 - y} z{\rm{d}}z{\rm{d}}y{\rm{d}}x = \frac{1}{3}$
(b) $yz$-plane
$\mathop \smallint \limits_{}^{} \mathop \smallint \limits_{}^{} \mathop \smallint \limits_{\cal W}^{} z{\rm{d}}V = \mathop \smallint \limits_{y = 0}^2 \mathop \smallint \limits_{z = 0}^{2 - y} \mathop \smallint \limits_{x = 0}^{y/2} z{\rm{d}}x{\rm{d}}z{\rm{d}}y = \frac{1}{3}$
(c) $xz$-plane
$\mathop \smallint \limits_{}^{} \mathop \smallint \limits_{}^{} \mathop \smallint \limits_{\cal W}^{} z{\rm{d}}V = \mathop \smallint \limits_{x = 0}^1 \mathop \smallint \limits_{z = 0}^{2 - 2x} \mathop \smallint \limits_{y = 2x}^{2 - z} z{\rm{d}}y{\rm{d}}z{\rm{d}}x = \frac{1}{3}$
The three answers agree.
Work Step by Step
We have the region ${\cal W}$ bounded by
$y+z=2$, ${\ \ \ }$ $2x=y$, ${\ \ \ }$ $x=0$, ${\ \ \ }$ and $z=0$
(a) Projecting ${\cal W}$ onto the $xy$-plane we obtain the domain ${\cal D}$, which is both vertically and horizontally simple region. From Figure 15 and the figure attached, we notice that ${\cal D}$ is bounded by the line $2x=y$. We choose to describe ${\cal D}$ as a vertically simple region, so the domain description is
${\cal D} = \left\{ {\left( {x,y} \right)|0 \le x \le 1,2x \le y \le 2} \right\}$
Thus, the region ${\cal W}$ consists of all points lying between ${\cal D}$ and the upper face $y+z=2$. That is, the $z$-coordinate satisfies $0 \le z \le 2 - y$. So, the region description of ${\cal W}$ is
${\cal W} = \left\{ {\left( {x,y,z} \right)|0 \le x \le 1,2x \le y \le 2,0 \le z \le 2 - y} \right\}$
So, the triple integral of $f\left( {x,y,z} \right) = z$ is
$\mathop \smallint \limits_{}^{} \mathop \smallint \limits_{}^{} \mathop \smallint \limits_{\cal W}^{} z{\rm{d}}V = \mathop \smallint \limits_{x = 0}^1 \mathop \smallint \limits_{y = 2x}^2 \mathop \smallint \limits_{z = 0}^{2 - y} z{\rm{d}}z{\rm{d}}y{\rm{d}}x$
$ = \frac{1}{2}\mathop \smallint \limits_{x = 0}^1 \mathop \smallint \limits_{y = 2x}^2 \left( {{z^2}|_0^{2 - y}} \right){\rm{d}}y{\rm{d}}x$
$ = \frac{1}{2}\mathop \smallint \limits_{x = 0}^1 \mathop \smallint \limits_{y = 2x}^2 \left( {4 - 4y + {y^2}} \right){\rm{d}}y{\rm{d}}x$
$ = \frac{1}{2}\mathop \smallint \limits_{x = 0}^1 \left( {\left( {4y - 2{y^2} + \frac{1}{3}{y^3}} \right)|_{2x}^2} \right){\rm{d}}x$
$ = \frac{1}{2}\mathop \smallint \limits_{x = 0}^1 \left( {\frac{8}{3} - 8x + 8{x^2} - \frac{8}{3}{x^3}} \right){\rm{d}}x$
$ = \frac{1}{2}\left( {\frac{8}{3}x - 4{x^2} + \frac{8}{3}{x^3} - \frac{2}{3}{x^4}} \right)|_0^1$
$ = \frac{1}{2}\left( {\frac{8}{3} - 4 + \frac{8}{3} - \frac{2}{3}} \right) = \frac{1}{3}$
So, $\mathop \smallint \limits_{}^{} \mathop \smallint \limits_{}^{} \mathop \smallint \limits_{\cal W}^{} z{\rm{d}}V = \mathop \smallint \limits_{x = 0}^1 \mathop \smallint \limits_{y = 2x}^2 \mathop \smallint \limits_{z = 0}^{2 - y} z{\rm{d}}z{\rm{d}}y{\rm{d}}x = \frac{1}{3}$.
(b) Projecting ${\cal W}$ onto the $yz$-plane we obtain the domain ${\cal T}$, which is both vertically and horizontally simple region. From Figure 15 and the figure attached, we notice that ${\cal T}$ is bounded by the line $z=2-y$. We choose to describe ${\cal T}$ as a vertically simple region, so the domain description is
${\cal T} = \left\{ {\left( {y,z} \right)|0 \le y \le 2,0 \le z \le 2 - y} \right\}$
Thus, the region ${\cal W}$ consists of all points lying between ${\cal T}$ and the left face $2x=y$. That is, the $x$-coordinate satisfies $0 \le x \le \frac{y}{2}$. So, the region description of ${\cal W}$ is
${\cal W} = \left\{ {\left( {x,y,z} \right)|0 \le y \le 2,0 \le z \le 2 - y,0 \le x \le \frac{y}{2}} \right\}$
So, the triple integral of $f\left( {x,y,z} \right) = z$ is
$\mathop \smallint \limits_{}^{} \mathop \smallint \limits_{}^{} \mathop \smallint \limits_{\cal W}^{} z{\rm{d}}V = \mathop \smallint \limits_{y = 0}^2 \mathop \smallint \limits_{z = 0}^{2 - y} \mathop \smallint \limits_{x = 0}^{y/2} z{\rm{d}}x{\rm{d}}z{\rm{d}}y$
$ = \mathop \smallint \limits_{y = 0}^2 \mathop \smallint \limits_{z = 0}^{2 - y} z\left( {x|_0^{y/2}} \right){\rm{d}}z{\rm{d}}y$
$ = \frac{1}{2}\mathop \smallint \limits_{y = 0}^2 \mathop \smallint \limits_{z = 0}^{2 - y} yz{\rm{d}}z{\rm{d}}y$
$ = \frac{1}{4}\mathop \smallint \limits_{y = 0}^2 y\left( {{z^2}|_0^{2 - y}} \right){\rm{d}}y$
$ = \frac{1}{4}\mathop \smallint \limits_{y = 0}^2 y\left( {4 - 4y + {y^2}} \right){\rm{d}}y$
$ = \frac{1}{4}\left( {2{y^2} - \frac{4}{3}{y^3} + \frac{1}{4}{y^4}} \right)|_0^2$
$ = \frac{1}{4}\left( {8 - \frac{{32}}{3} + 4} \right)$
So, $\mathop \smallint \limits_{}^{} \mathop \smallint \limits_{}^{} \mathop \smallint \limits_{\cal W}^{} z{\rm{d}}V = \mathop \smallint \limits_{y = 0}^2 \mathop \smallint \limits_{z = 0}^{2 - y} \mathop \smallint \limits_{x = 0}^{y/2} z{\rm{d}}x{\rm{d}}z{\rm{d}}y = \frac{1}{3}$.
(c) Projecting ${\cal W}$ onto the $xz$-plane we obtain the domain ${\cal S}$, which is both vertically and horizontally simple region. From Figure 15 and the figure attached, we notice that ${\cal S}$ is bounded by the line $z=2-2x$. We choose to describe ${\cal S}$ as a vertically simple region, so the domain description is
${\cal S} = \left\{ {\left( {x,z} \right)|0 \le x \le 1,0 \le z \le 2 - 2x} \right\}$
We see from the Figure 15, ${\cal W}$ is bounded below by the left face $2x=y$ and bounded above by the upper face $y+z=2$. Thus, the $y$-coordinate satisfies $2x \le y \le 2 - z$. So, the region description of ${\cal W}$ is
${\cal W} = \left\{ {\left( {x,y,z} \right)|0 \le x \le 1,0 \le z \le 2 - 2x,2x \le y \le 2 - z} \right\}$
So, the triple integral of $f\left( {x,y,z} \right) = z$ is
$\mathop \smallint \limits_{}^{} \mathop \smallint \limits_{}^{} \mathop \smallint \limits_{\cal W}^{} z{\rm{d}}V = \mathop \smallint \limits_{x = 0}^1 \mathop \smallint \limits_{z = 0}^{2 - 2x} \mathop \smallint \limits_{y = 2x}^{2 - z} z{\rm{d}}y{\rm{d}}z{\rm{d}}x$
$ = \mathop \smallint \limits_{x = 0}^1 \mathop \smallint \limits_{z = 0}^{2 - 2x} z\left( {y|_{2x}^{2 - z}} \right){\rm{d}}z{\rm{d}}x$
$ = \mathop \smallint \limits_{x = 0}^1 \mathop \smallint \limits_{z = 0}^{2 - 2x} z\left( {2 - z - 2x} \right){\rm{d}}z{\rm{d}}x$
$ = \mathop \smallint \limits_{x = 0}^1 \mathop \smallint \limits_{z = 0}^{2 - 2x} \left( {2z - {z^2} - 2xz} \right){\rm{d}}z{\rm{d}}x$
$ = \mathop \smallint \limits_{x = 0}^1 \left( {\left( {{z^2} - \frac{1}{3}{z^3} - x{z^2}} \right)|_0^{2 - 2x}} \right){\rm{d}}x$
$ = \mathop \smallint \limits_{x = 0}^1 \left( {{{\left( {2 - 2x} \right)}^2} - \frac{1}{3}{{\left( {2 - 2x} \right)}^3} - x{{\left( {2 - 2x} \right)}^2}} \right){\rm{d}}x$
$ = \mathop \smallint \limits_{x = 0}^1 \left( { - \frac{4}{3}{x^3} + 4{x^2} - 4x + \frac{4}{3}} \right){\rm{d}}x$
$ = \left( { - \frac{1}{3}{x^4} + \frac{4}{3}{x^3} - 2{x^2} + \frac{4}{3}x} \right)|_0^1$
$ = - \frac{1}{3} + \frac{4}{3} - 2 + \frac{4}{3} = \frac{1}{3}$
So, $\mathop \smallint \limits_{}^{} \mathop \smallint \limits_{}^{} \mathop \smallint \limits_{\cal W}^{} z{\rm{d}}V = \mathop \smallint \limits_{x = 0}^1 \mathop \smallint \limits_{z = 0}^{2 - 2x} \mathop \smallint \limits_{y = 2x}^{2 - z} z{\rm{d}}y{\rm{d}}z{\rm{d}}x = \frac{1}{3}$.
The three answers agree.
